Fixed-point iteration 'show that' exercise

Hi,

problem:

The iteration defined by , where ,
has two fixed points where . Show that

and deduce that . How does the iteration behave for other values of ?

attempt at solution:

and since,

we have that,

I am a bit confused regarding the limit-question. If and it means that could be any number larger than 1 but smaller than . I figure that I could then choose some such that , in other words we would be moving away from .

Now if instead of , I could see how .

As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

The iteration defined by , where ,
has two fixed points where . Show that

and deduce that . How does the iteration behave for other values of ?

attempt at solution:

and since,

we have that,

I am a bit confused regarding the limit-question. If and it means that could be any number larger than 1 but smaller than . I figure that I could then choose some such that , in other words we would be moving away from .

Now if instead of , I could see how .

As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

If I now look at what happens if in a similar fashion, I get that, and so the sequence will not converge to .

I think this is because the derivative of is and when it means that and so the derivative is larger than 1. If our function has a derivative larger than 1 in the interval we are working on, fixed-point iteration will not converge.

The iteration defined by , where ,
has two fixed points where . Show that
...

As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

Thanks.

May be you are looking for this.
I shall denote the sequence by for and define for
where .
It is clear that for , .
Now, for , compute................_
Let for .
Repeating the procedure above with the abbreviation that the empty product
is the unity, we get for all that................_................_
which implies that the sequence is either non-decreasing or non-increasing.
Therefore, exists (finite or infinite).
Taking as on both sides of (1), we get
which gives the equlibrium points by solving the quadratic equation and for .
If satisfies (3), then , and using (1), we have for all .
As the term in (2) is quadratic, we can solve for and get that and for .

If , then , and the sequence is non-increasing.
Then the sequence can not converge a value greater than .
Hence, if , then .

Now, let , then , and the sequence is strictly increasing.
By induction, we have for all implying that .

This shows that the equilibrium point is stable.

In the case, , we have for all , i.e., .

Finally, if , then , and the sequence is strictly increasing.
By induction, we have for all implying that .

thank you for what looks like a wonderful answer. The only problem is that I do not understand it. First of all, some of the notation is giving me problems.
You first denote the sequence by .
I think .
You then write . I have not seen this notation before, with the sign and everything, but from the way is defined in the problem, I'm assuming that it means ?

Again, thanks for taking the time to write out all that good stuff. I just hope it isn't wasted on me. Will try to understand it!

thank you for what looks like a wonderful answer. The only problem is that I do not understand it. First of all, some of the notation is giving me problems.
You first denote the sequence by .
I think .
You then write . I have not seen this notation before, with the sign and everything, but from the way is defined in the problem, I'm assuming that it means ?

Again, thanks for taking the time to write out all that good stuff. I just hope it isn't wasted on me. Will try to understand it!

Hi Mollier,

and
I am glad you like the solution.
Actually, what I have shown you above is the standard procedure to be followed for such kind of problems.

Now, for , compute................_
Let for .
Repeating the procedure above with the abbreviation that the empty product
is the unity, we get for all that................_................_
which implies that the sequence is either non-decreasing or non-increasing.
Therefore, exists (finite or infinite).

Hi again bkarpuz,

thanks for the quick intro to set-theory notation

I can see how you get,

,

which you then write as,

.

If I am reading this right, the sequence is decreasing if,

,

and if I'm doing this right then,

.

Then has one more term than and is therefore bigger.

You rewrite in several different ways and then say that the sequence is non-decreasing or non-increasing. I do not get that though..

Up to now everything is okay.
But do not forget that we do not yet know how large is the product, i.e., or .

Originally Posted by Mollier

If I am reading this right, the sequence is decreasing if,

,

A sequence is decreasing if each term is greater than preceding one (or succeeding terms are greater than the present one), i.e., (or ) for all .
And conversely, increasing if .
Actually, the forward difference operator notation illustrates this better (analogue to differential operator).
With this notation
1. A sequence is increasing if for all . For instance, .
2. A sequence is decreasing if for all . For instance, .
3. A sequence is nondecreasing if for all . For instance, , where is the least integer function.
4. A sequence is nonincreasing if for all .
In each case, the sequence is monotonic, and it has a limit. (compare this with derivative).

Originally Posted by Mollier

and if I'm doing this right then,

.

Then has one more term than and is therefore bigger.

Also I can say that
5. If ( ), which is equivalent to ( ), then the sequence is convex. For instance, .
If an increasing sequence is convex, then it tends to infinity.
6. If , which is equivalent to ), then the sequence is concave.

Including more terms in the product on the right hand side does not mean that the sequence is decreasing or increasing.
For instance, , which gives .